Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
  1. Expand in Fourier series of $f(x) = \sin x$ for $0<x<l$. Deduce the result \[ \frac1{1 \cdot 3} - \frac{1}{3\cdot5} +\frac{1}{5\cdot 7} - \cdots = \pi-\frac{2}{4}. \]

  2. Obtain half range sine series for $f(x)$ in $0<x<l$ and deduce the series \[\sum\frac{1}{n^2} = \frac{{\pi^2}}6.\]

share|cite|improve this question
Perhaps you should explain what you have tried already? – copper.hat Jul 20 '12 at 19:01
I tried to fix the source and tags — whatever you typed wasn't displaying at all on my end, so I tried to put everything in LaTeX. Please check for errors. [It's a good idea to try to learn the rudiments of LaTeX for the purpose of asking questions here, by the way.] In particular, I couldn't tell what you were trying to write in that first series. It doesn't look right to me as it is. – Dylan Moreland Jul 20 '12 at 19:02
Presumably, the term $\frac 1 {3\cdot 7}$ is meant to be $\frac 1 {5\cdot 7}$, or else I don't see the pattern to the series. – Thomas Andrews Jul 20 '12 at 19:05
@ThomasAndrews Change made. – Dylan Moreland Jul 20 '12 at 19:22
@ThomasAndrews I hv the same question as it is written....May be there is a printing mistake...Please can you provide me the steps of the solution taking 1/5.7 if you think its correct....and the steps for the second part too....please.... – Naman Sindhi Jul 20 '12 at 20:08

Hint :

Expand the periodic function

$$f(x) = \left\{ \begin{array}{rcl} {-k} & \mbox {when} & -\pi \lt x \lt 0 \\ +k & \mbox{when} & 0 \lt x \lt \pi \end{array}\right.$$ and when $f(x+2\pi)=f(x)$.

To obtain the Fourier coefficients $a_n$ and $b_n$ you do the following integration $$a_n= \frac{1}{\pi}\int_{-\pi}^{+\pi}{k\cos(nx)dx}$$ and $$b_n= \frac{1}{\pi}\int_{-\pi}^{+\pi}{k\sin(nx)dx}$$ This will show that $a_n=0$ and $b_n=\frac{4k}{n\pi}$ when n is odd and $b_n=0$ when n is even. So $$f(x)=\frac{4k}{\pi}\left(\sin x + \frac{1}{3}\sin 3x+\frac{1}{5}\sin 5x+\dots\right)$$ Now we know $f(x)=+k$ when $ 0 \lt x \lt \pi$. So $f(\frac{\pi}{2})=+k$ and hence $$ k=\frac{4k}{\pi}\left(\sin\frac{\pi}{2} + \frac{1}{3}\sin3\frac{\pi}{2}+\frac{1}{5}\sin5\frac{\pi}{2}+\dots\right)$$ $$\Rightarrow \frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots$$ Now let $$ \frac{1}{1 \cdot 3} - \frac{1}{3\cdot5} +\frac{1}{5\cdot 7} - \cdots = S_1. $$$$\Rightarrow S_1=\frac{1}{2}\left[\frac{3-1}{1 \cdot 3}- \frac{5-3}{3\cdot5} +\frac{7-5}{5\cdot 7} - \cdots\right]$$ $$\Rightarrow 2S_1=\left[1-2\cdot\frac{1}{3}+2\cdot\frac{1}{5}-2\cdot\frac{1}{7}- \cdots\right]$$$$\Rightarrow 2S_1=\left[1+2\left(\frac{\pi}{4}-1\right)\right]$$$$\Rightarrow S_1=\frac{\pi}{4}-\frac{1}{2}$$$$\Rightarrow \frac{1}{1 \cdot 3} - \frac{1}{3\cdot5} +\frac{1}{5\cdot 7} - \cdots=\frac{\pi}{4}-\frac{1}{2}$$ I think this is what you are looking for in the first part.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.